Answer:
percentage by mass of Ba = 0.4583/0.8360 × 100 = 45.83
/0.8360 = 54.82%
Explanation:
The mass of Barium in the original compound is the same as the masss of barium in the precipitate BaSO4.
Ba²+
molar mass of BaSO4 = 137 + 32 + 16(4) = 137 + 32 + 64 = 233 g/ mol
since
137 grams of barium is in 233 g of BaSO4
? grams of barium is in 0.7794 g of BaSO4
cross multiply
grams of barium = 0.7794 × 137/233
grams of barium = 106.7778/233
grams of barium = 0.45827381974
grams of barium = 0.4583 grams
percentage by mass of Barium in the original compound =
mass of barium/ mass of the compound × 100
mass of barium = 0.4583 g
mass of compound = 0.8360 g
percentage by mass of Ba = 0.4583/0.8360 × 100 = 45.83
/0.8360 = 54.82%
